Free Objective Test 02 Practice Test - 11th and 12th

Question 1

The earth goes around the sun with a fixed time period due to the gravitational force of attraction between them.

Here the force of gravity (Fg) on earth is the restoring force.

True

False

SOLUTION

Solution : B

So let's try to find equilibrium. But clearly, there is none.

At every point the earth is acted upon by Sun's gravitational pull.

Question 2

Which of these situations shown are in stable equilibrium?

SOLUTION

Solution : A, C, and D

Stable equilibrium is the position where the particle has lowest potential energy.

If you displace the particle from this position then it will have a tendency to come back to the original position as a restoring force would have started acting on it.

If I displace the ball slightly to left or right it will have a tendency to go towards its equilibrium position.

Also the ball has minimum potential energy there.

Stable equilibrium

The ball has highest potential energy at this position. If displaced slightly will fall away from the starting point.
So Unstable Equilibrium

Since spring in natural length so potential energy is 0.

Also if the block is displaced little to the left, the left spring will get compressed and push to the right, towards the starting position and at the same time the right spring will get elongated and pull towards the right i.e., towards the starting point.

So Stable equilibrium

The wood half submerged and half floating. Its weight is equal to the buoyant force water applies on it.
If I push it little more in water the buoyant force will increase and try to push it up towards the starting position. If I try to pull it out a little it has a tendency to go back down to the initial position because of its own weight.

Stable Equilibrium

Question 3

When a body repeats its motion in regular interval of time then it's said to execute periodic motion. Let's say a car is going on a race track (as shown in figure)

The car covers 5 laps.

For each lap it took 10 mins

In which of the below mentioned cases will its motion be considered periodic.

Case I: The car was travelling with constant speed.

Case II: The car had a constant acceleration from rest (0 m/s) and then a constant deceleration to rest (0 m/s) in each lap for all 5 laps.

Case III: The car had a non-uniform acceleration throughout the journey.

Case IV: lap 1 with constant speed, lap 2 with constant acceleration, lap 3, 4, 5 with variable acceleration.

Case I is periodic

Case II is periodic

Case III is Periodic

All cases are periodic

SOLUTION

Solution : A and B

You know that a motion that repeats itself in equal intervals of time is called periodic, right?

Let's discuss what it means exactly,

It means that after some specific interval of time T the position of the body from the reference point and its velocity become equal to what they were at the reference point. so no matter

what reference point you choose in the path, after time T the motion "repeats” itself.

In case I and II if we choose any reference point in the path we find after passing of time equivalent to a multiple of T=10 min the motion "repeats” itself.

This wouldn't happen for case III and IV.

Question 4

A person goes to bed at sharp 10.00 pm every day. Is it an example of periodic motion?

True

False

SOLUTION

Solution : B

Here, the person is going to bed sharp at 10:00.

But what about his motion, if we take his bed as the reference he might take different paths every day. Someday he might follow.

(1) Bed - Bathroom - Bus stop - Office - Bus stop - Bed

(2) Bed - Bathroom - Cab - Office - Restaurant - Bed

So every day the path might be different, even if the path is same the velocity might be different (at the same time on different days)

Example: He might go to his office every day but the speed of a bike and a car and a bus will be different.

Clearly, the motion is not periodic.

Question 5

A particle is performing a linear simple harmonic motion. If the acceleration and the corresponding velocity of the particle are a and v respectively, which of the following graphs is correct?

SOLUTION

Solution : C

If a particle performs SHM with angular frequency ' $ω$ ' and amplitude 'A', then its displacement from mean position will be equal to $x = A s i n ω t$ , if its initial phase is equal to zero.

Velocity will be equal to

$v = \frac{d x}{d t} = A ω c o s ω t \to (1)$ and acceleration will be $α = \frac{d v}{d t} = - A ω^{2} s i n (ω t) \to$ (2)

From these equations, $c o s ω t = \frac{v}{A ω}, s i n ω t = \frac{α}{- A ω^{2}}$

Squaring and adding, $\frac{v^{2}}{A^{2} ω^{2}} + \frac{α^{2}}{A^{2} ω^{4}} = 1$

If v is taken on the y-axis and $α$ on the x-axis, then it will be an ellipse

Question 6

A particle having mass 10g oscillates according to the equation $x = (2.0 c m) s i n [(100 s^{- 1}) t + \frac{π}{6}]$ . Find the spring constant

$200 k g / s^{2}$

$100 k g / s^{2}$

$10, 000 k g / s^{2}$

Data insufficient

SOLUTION

Solution : B

A general sinusoidal, oscillation of amplitude A, angular frequency $ω a n d a t a p h a s e ψ$ is given as

$x (t) = A s i n (ω t + ψ)$ ,

Compare this with the equation under discussion

$x = [(0.2 c m) s i n (100 s^{- 1}) t + \frac{π}{6}]$

It is clear from a comparison, that

$A = 0.2 c m = 0.002 m$

$ω = 100 s^{- 1}, a n d ψ = \frac{π}{6}$

We know that for oscillations in a spring mass system, the spring constant is related to the mass m and ω as

$k = m ω^{2}$

The mass is given to us, viz. $m = 10 g = 0.01 k g$

$∴ T h e k = [0.01 \times 1002] k g s^{- 2}$

$= 100 k g s^{- 2}$

Hence correct option is (b).

Question 7

A small creatures moves with constant speed in a small vertical circle on a bright day. Does its shadow formed by the sun on a horizontal plane move in a simple harmonic motion?

yes, only if its noon

yes, at any time of the day

No, the motion will be harmonic but not SHM

Data insufficient

SOLUTION

Solution : B

Well! When the sun is exactly above the vertical circle, the shadow is formed and it does SHM very much as we know from the example

in the video linking SHM to circular motion. Question here actually is "will the shadow's motion be a SHM when the sun is at some angle (say $θ$ ) ?”

Let's explore that

Let's say that the sun light from the sun makes an angle θ with the horizontal as you can see is the diagram. Since the creature moves with constant speed, the time it takes to move from A to B is the same it takes to move from B to C. Also a little geometry shows us that AB = BC. The time taken to move from C to D and then again from D to A is again same. The time period of the SHM of the shadow is castant.

Now let's check for it's acceleration at some point P after a time 't' has elapsed.

The horizontal component of acceleration at $P i s a_{C h}$

$a_{C h} = a s i n (ω t + \frac{π}{2} - θ)$

which is an acceleration relation for SHM.

Question 8

Figure shows three identical springs A, B, C. when a 4 kg weight is hung on A, it descends by 1 cm. When a 6kg weight is hung on C, it will descend by

A. 1.5 cm

B. 3.0 cm

C. 4.5 cm

D. 6.0 cm

SOLUTION

Solution : B

Let k be the force constant of each spring. The force F needed to extend the spring A by an amount x is
$F = k x o r x = \frac{F}{k}$
In Fig, the springs are connected in series. Therefore, the force constant k’ of the combination is given by
$\frac{1}{k^{'}} = \frac{1}{k} + \frac{1}{k}$
Or k' = $\frac{k}{2}$
If a force F’ is applied, the combination will extend by
$x^{'} = \frac{F^{'}}{k^{'}}$
Thus $\frac{x^{'}}{x}$ = $\frac{F^{'}}{F} . \frac{k}{k^{'}}$ = $\frac{6}{4} \times \frac{k}{k / 2}$ =3
Or x' = 3x = $3 \times 1 c m = 3 c m$
Hence the correct choice is (b).

Question 9

When a mass m is hung from the lower of a spring of negligible mass, an extension x is produced in the spring. The mass is set into vertical oscillations.The time period of oscillation is

T = 2 π \sqrt{\frac{x}{m g}}

T = 2 π \sqrt{\frac{g x}{m}}

T = 2 π \sqrt{\frac{x}{g}}

T = 2 π \sqrt{\frac{x}{2 g}}

SOLUTION

Solution : C

If k is the force constant, we have
mg=kx
or $\frac{m}{k} = \frac{x}{g}$
$∴ T = 2 π \sqrt{\frac{m}{k}} = 2 π \sqrt{\frac{x}{g}}$

Question 10

A spring with no mass attached to it hangs from a rigid support. A mass m is now hung on the lower end to the spring. The mass is supported on a platform so that the spring remains relaxed. The supporting platformis then suddenly removed and the mass begins to oscillate. The lowest position of the mass during the oscillation is 5 cm below the place where it was resting on the platform. What is the angular frequency of oscillation? Take g = 10 $m s^{- 2}$

10 r a d s^{- 1}

20 r a d s^{- 1}

30 r a d s^{- 1}

40 r a d s^{- 1}

SOLUTION

Solution : B

It is clear that the separation between the two extreme positions of the oscillating mass is 5 cm. Therefore, the equilibrium position 0 is 2.5 cm below the supporting platform. In other words, force mg produces an extension y = 2.5 cm in the spring. If k is the force constant of the spring we have
m g=k y
or $\frac{k}{m} = \frac{g}{y} = \frac{1000 c m s^{- 2}}{2.5 c m} = 400 s^{- 1}$
The angular frequency $ω$ of oscillation is
$ω = \sqrt{\frac{k}{m}} = \sqrt{400} = 20 r a d s^{- 1}$

Question 11

In a spring block system the block is displaced by a distance of A from its mean position and left to undergo Simple harmonic motion. The displacement of the block in one time period is:

A. A

B. 2A

C. 4A

D. Zero

SOLUTION

Solution : D

In one time period the particle comes back to its starting position, irrespective of from where did it start. So the displacement has to be 0.

Question 12

A spring block system is executing SHM with angular frequency of 1 rad/sec and has an amplitude of 2m. Find the maximum kinetic energy of the system.Given mass of the block is 2 kg.

A. 1J

B. 4J

C. 10J

D. Data insufficient

SOLUTION

Solution : B

$K E = \frac{1}{2} m (ω \sqrt{A^{2} - x^{2}})^{2}$
K.E. is max at x = 0 (mean position)
$\Rightarrow K E_{m a x} = \frac{1}{2} m (ω^{2} A^{2})$
$= \frac{1}{2} 2 (2^{2})$
$K E_{m a x} = 4 J .$

Question 13

A particle executes simple harmonic motion with a frequency. f. The frequency with which its kinetic energy oscillates is

A. f/2

B. f

C. 2f

D. 4f

SOLUTION

Solution : C

During one complete oscillation, the kinetic energy will become maximum twice. Therefore the frequency of kinetic energy will be 2f.

Question 14

The time taken by a particle executing simple harmonic motion of time period T to move from the mean position to half the maximum displacement is

\frac{T}{2}

\frac{T}{4}

\frac{T}{8}

\frac{T}{12}

SOLUTION

Solution : D

Let the displacement of the particle be given by x = A sin $ω$ t = A sin $(\frac{2 π t}{T})$ i.e., when x = 0, $t_{0} = 0.$
When x = A/2, the value of t is given by $\frac{A}{2} = A s i n \frac{2 π t_{1}}{T}; s i n \frac{2 π t_{1}}{T} = \frac{1}{2}$
$\frac{2 π t_{1}}{T} = \frac{π}{6} o r t_{1} = \frac{T}{12}; t_{1} - t_{0} = \frac{T}{12} - 0 = \frac{T}{12}$

Free Objective Test 02 Practice Test - 11th and 12th

Question 1

SOLUTION

Question 2

SOLUTION

Question 3

SOLUTION

Question 4

SOLUTION

Question 5

SOLUTION

Question 6

SOLUTION

Question 7

SOLUTION

Question 8

SOLUTION

Question 9

SOLUTION

Question 10

SOLUTION

Question 11

SOLUTION

Question 12

SOLUTION

Question 13

SOLUTION

Question 14

SOLUTION

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